【LeetCode】2.Array and String — Largest Number At Least Twice of Others 最大数至少次大数二倍

In a given integer array nums, there is always exactly one largest element.

Find whether the largest element in the array is at least twice as much as every other number in the array.

If it is, return the index of the largest element, otherwise return -1.

Example 1:

Input: nums = [3, 6, 1, 0]
Output: 1
Explanation: 6 is the largest integer, and for every other number in the array x,
6 is more than twice as big as x.  The index of value 6 is 1, so we return 1.

Example 2:

Input: nums = [1, 2, 3, 4]
Output: -1
Explanation: 4 isn‘t at least as big as twice the value of 3, so we return -1.

Note:

1. nums will have a length in the range [1, 50].
2. Every nums[i] will be an integer in the range [0, 99].

题目简单明了，寻找数组总最大的数和次大的数然后按要求比较即可。

class Solution {
public:
    int dominantIndex(vector<int>& nums) {
        if(nums.size()==0) return -1;
        if(nums.size()==1) return 0;
        int Max=nums[0];
        int NextMax=0;
        int index=0;
        //寻找最大的数和次大的数
        for(auto i = 1; i<nums.size();i++)
        {
            if(Max < nums[i])
            {
                NextMax=Max;
                Max = nums[i];
                index = i;
            }
            else if(NextMax <= nums[i])
            {
                NextMax = nums[i];
            }
        }
        if(Max >= 2*NextMax) return index;
        return -1;   
    }
};

【LeetCode】2.Array and String — Largest Number At Least Twice of Others 最大数至少次大数二倍

原文：https://www.cnblogs.com/hu-19941213/p/10930154.html